Question: Factor completely. $81-4x^2=$
Both $81$ and $4x^2$ are perfect squares, since $81=({9})^2$ and $4x^2=({2x})^2$. $81-4x^2 = ({9})^2-({2x})^2$ So we can use the difference of squares pattern to factor. ${a}^2 - {b}^2 =({a}+{b})({a}-{b})$ In this case, ${a}={9}$ and ${b}={2x}$ : $({9})^2 - ({2x})^2 =({9}+{2x})({9}-{2x})$ In conclusion, $81-4x^2=(9+2x)(9-2x)$ Remember that you can always check your factorization by expanding it.